The difference between the two acute angles of a right-angled triangle is

Given:

Difference between two acute angles of a right-angled triangle $=\frac{2 \pi}{5} \mathrm{rad}$

$\because 1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{\circ}$

$\therefore \frac{2 \pi}{5} \mathrm{rad}=\left(\frac{180}{\pi} \times \frac{2 \pi}{5}\right)^{\circ}$

$=(36 \times 2)^{\circ}$

$=72^{\circ}$

Now, let one acute angle of the triangle be $x^{\circ}$.

Therefore, the other acute angle will be $90^{\circ}-x^{\circ}$.

Now,

$x^{\circ}-\left(90^{\circ}-x^{\circ}\right)=72^{\circ}$

$\Rightarrow x-90+x=72$

$\Rightarrow 2 x=162$

$\Rightarrow x=81$

Thus, we have:

$x^{\circ}=81^{\circ}$

And,

$90^{\circ}-x^{\circ}=90^{\circ}-81^{\circ}=9^{\circ}$